Alternate Design Method- Design Procedure of Two-way Slabs using ACI Moment Coefficients and Approved by BNBC 2013

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1 Alternate Design Method- Design Procedure of Two-way Slabs using ACI Moment Coefficients and Approved by BNBC 2013 Scope and Limitations (BNBC) The provisions of this section may be used as alternative to those of through for two way slabs supported on all four edges by walls, steel beams or monolithic concrete beams having a total depth not less than 3 times the slab thickness Panels shall be rectangular with a ratio of longer to shorter span centre to centre of supports not greater than 2. Step 01: Determination of thickness of the slab panel. Determine the thickness of the slab panel using previous article. Step 02: Calculation of factored load. W u = 1.2*DL+1.6*LL where DL= Total dead load (i.e.: Slab self weight, Floor finish, Partition wall, Plaster etc.) LL= Live load. Step 03: Determination of moment coefficients. m= A B where A= Shorter length of the slab. B= Longer length of the slab. Case type is identified from end condition. Using the value of m corresponding moment coefficients are obtained for respective case type from corresponding tables. The co-efficients are: C A neg and C B neg C A DL pos and C B DL pos C A LL pos and C B LL pos Page 21

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5 Step 04: Calculation of moments. Positive moments: +M A = C A DL W DL L A 2 + C A LL W LL L A 2 ; +M B = C B DL W DL L B 2 + C B LL W LL L B 2. Negative Moments: -M A = C A, neg W T L A 2 ; -M B = C A, neg W T L B 2 ; W LL = Uniform Live load per unit area= 1.6 LL, (Factored since USD will be used) W DL = Uniform Dead load per unit area = 1.2 DL (Factored since USD will be used) W u =W T = Total Uniform load per unit area = 1.2*DL+1.6*LL (Factored since USD will be used) Start with Max moment, M then, M A s =.9 fy (d a 2 ) Now, find a = As fy 0.85 f c b Then, do at least another trial, with new a, and find area of steel. Last Updated: May 13, 2017 The following Example was done by Md. Mahmudun Nobe, ID , AUST Batch no. 28 Design of two-way edge supported slab by using moment coefficients. Beam-column supported floor slab of a 93ꞌ-6ꞌꞌ 75ꞌ-6ꞌꞌ (center to center distance of extreme columns) cyclone shelter is to carry service live load of 100 psf in addition to its own weight, 1/2ꞌꞌ thick plaster and 3/2ꞌꞌ thick floor finish. Supporting columns of 14 in square are spaced orthogonally at an interval at 31ꞌ-2ꞌꞌ and 25ꞌ-2ꞌꞌ on centers along longitudinal and transverse directions respectively. Width of each beam is 14 in. Using BNBC/ACI code of moment coefficients design the slab by USD method, if fꞌ c = 3000 psi and f y = psi. Page 25

6 Solution: 31ꞌ-2ꞌꞌ = 93ꞌ-6ꞌꞌ 25ꞌ-2ꞌꞌ = 75ꞌ-6ꞌ Figure 03: Slab panel orientation and case type, e.g., case 9 is typical exterior, 4 is corner slab etc. Here A= 25ꞌ2ꞌꞌ-1ꞌ2ꞌꞌ= 24ꞌ and B= 31ꞌ2ꞌꞌ-1ꞌ2ꞌꞌ= 30ꞌ= l n. t= l n (0.8+ fy ) ( = 36+9β = 8.38ꞌꞌ 8.5ꞌꞌ say. 24 So d= 8.5ꞌꞌ-1ꞌꞌ= 7.5ꞌ W DL = ( )*12.5*1.2= psf W LL = 100*1.6= 160 psf W u = psf m= A/B= 24/30= C A C B C A DL C B DL C A LL C B LL Controlling coefficient. [Note: In this slab, there are four different types of cases among all panels. We take the maximum value of moment coefficient from four cases.] +M A = C A DL *W DL *A 2 + C A LL *W LL *A 2 = 0.039*157.5* *160*24 2 = l lb-ft/ft = 7.96 k-ft/ft -M A = C A *W u *A 2 = 0.075*317.5*24 2 = lb-ft/ft = 13.6 k-ft/ft +M B = C B DL *W DL *B 2 + C B LL *W LL *B 2 = 0.016*157.5* *160*30 2 = 5148 lb-ft/ft = k-ft/ft -M B = C B *W u *B 2 Page 26

7 = 0.041*317.5*30 2 = lb-ft/ft = k-ft/ft Rebar for short direction/transverse direction: M 12 M 12 M A S A = = = = = in2 /ft (Controlling). and a = 0.9 f y d a d a d a 2 A s f y 0.85f c b = A s = 1.96*A S = 1.96*0.244 = in. A min = xbxt = x12x8.5 = in 2 /ft Using φ10mm bar Area of bar used width of strip S = = = 5.95ꞌꞌ 5.5ꞌꞌc/c at bottom along short direction Requried A s crank 50% bar to negative zone. M A S A = = = in2 /ft (Controlling). 4.5 d a 2 a = 1.96 A s = in A min = in 2 /ft. Already provided A s1 = = in 2 /ft 11 Extra top required, A s2 = ( ) = in 2 /ft. Using Φ10mm bar S= ꞌꞌ c/c extra top. Rebar along long direction: +A S B = = in2 /ft A min = in 2 /ft (Controlling). Using Φ10 mm c/c at bottom along long direction crank 50% bar to negative zone. -A S B = = in2 /ft Already provided A s1 = = in 2 /ft Extra top required, A s2 = ( ) in 2 /ft = in 2 /ft Using Φ 10mm ꞌꞌ c/c extra top. Corner Reinforcement (BNBC 2013) a) Corner reinforcement shall be provided at exterior corners in both bottom and top of the slab, for a distance in each direction from the corner equal to one fifth the longer span of the corner panel as per provisions of Corner reinforcement in both top and bottom of slab shall be sufficient to resist a moment per unit of width equal to the maximum positive moment per unit width in the slab panel The moment shall be assumed to be about an axis perpendicular to the diagonal from the corner in the top of the slab and about an axis parallel to the diagonal from the corner in the bottom of the slab Corner reinforcement shall be provided for a distance in each direction from the corner equal to one fifth the longer span Corner reinforcement shall be placed parallel to the diagonal in the top of the slab and perpendicular to the diagonal in the bottom of the slab. Alternatively, reinforcement shall be placed in two layers parallel to the sides of the slab in both the top and bottom of the slab. Page 27

8 Figure: Reinforcement details of slab in plan. Page 28